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```An Introduction to Heat Transport by Conduction
and Fourier's Law

By Patrick Bruskiewich

Fourier's Law

The relationship between the heat flow and the temperature difference was discovered
experimentally in 1804 by the scientist French scientist Biot, who discovered
experimentally that the driving force for heat was the temperature difference AT across
an object

q x cc -AT

The direction of the heat flow is from the higher temperature to the lower temperature
(refer to Fig. 1: The Direction of Heat Flow).

J

r a < r £

Fig. 1: The Direction of Heat Flow

The full theory of heat transport by conduction was explained by the French scientist Jean
Baptiste Joseph Fourier (1768-1830) and first presented to the world in 1812. In the
1812 French Science Prize competition the mathematicians Laplace, Lagrange and
Legendre assigned the problem to French scientists to discover "The Mathematical
Theory of the laws of the propagation of heat and the comparison of the results of this
theory with exact experiment."

Fourier would win the 1812 French Science Prize by deriving the expression which now
carries his name.

The full theory was published in book form in 1822 by Fourier in the book Theorie
Analytic de la Chaleur which is considered on par with Newton's book Principia
Mathematica. Theorie Analytic de la Chaleur is considered one of the Great Books of the
Western World and even today, nearly two centuries after its publication, a book worth

Heat Transport by conduction is described by Fourier's Law, which in one dimension is
given by the expression

AU 7 A AT

q x = — = - k A —

At Ax

where q x is the rate of heat flow (thermal energy AU per time interval A^), and where k

W

is the thermal conductivity (in units of ), A x is the area perpendicular to the heat

mK

flow, and where AT is the temperature difference across the material of width Ax .

The temperature difference across the material can be measured either in degrees Kelvin
or Celsius, since they are related one to the other by an additive value of 273 K. By
practice, an engineer or technologists would use the familiar Celsius scale, while a
scientists would tend to use the Kelvin scale.

The values for the thermal conductivity of different materials is determined through
experimental measurement and can be found in any good handbook or textbook. Some
representative examples for solid materials is given in Table 1 , Thermal Conductivity of
Solid Materials.

Material

W
k ( )
mK

Pure diamond

3400

Silver

427

Copper

397

Gold

314

Aluminium

238

Iron

79.5

34.7

Ice

2.0

Silica Brick

1.1

Standard Window Glass

0.84

Glass (pyrex)

0.8

Concrete

0.8

Compacted Snow

0.2

Wood

0.08

Loose Snow

0.05

Insulating Fibre Glass

0.035

Still Dry Air

0.0234

Table 1, Thermal Conductivity of Solid Materials

You will note that a good thermal insulator is also a good electrical insulator. There is a
theory that relates electron conductivity to thermal conductivity in good conductors that
explains this, that you can search for and read about if you are at all interested.

Example One: Conductive Heat Loss Through a Single
Window Pane of Glass

A simple example: of thermal conductivity is the conductive heat loss through a single
window pane of glass with an area A = 1.0 m 2 , with a thickness of Ax = 0.005 m . For

W

standard window glass k lass = 0.84 . Let the interior building temperature Jj be 22

mK

C, and the exterior building temperature T 2 be C.

A schematic of this system is given in Fig. 2. Conductive Heat Loss Through a Single
Window Pane of Glass.

T.=22C

7\ = CC

2a. = C.CC5m

Fig. 2. Conductive Heat Loss Through a Single Window Pane of Glass

The Temperature difference is then

AT = T 2 -T l =(0-22)C

By Fourier's law the conductive heat loss through this single window pane of glass is

AT

q x = -H

Ax

W ( 2 x(0-22)

= -0.84 1.0m 2 p } -

mK v } 0.005 m

= 3,696 W

which is a substantial amount of conductive heat that is leaving the inside of the building
and is lost to the surrounding air (in comparison the radiative heat loss is around 74.6
W, only 0.02 that of conductive heat loss).

Example Two: The Igloo as a Practical Winter Shelter

A second example of conductive heat loss is the simple Eskimo winter shelter made from
compacted snow known as the igloo. In the classical 1922 film Nanook of the North you
can watch an Eskimo family construct an igloo to protect against a night time temperature
of-25C.

The mean basal metabolism rate for a male adult Eskimo is around 10,000 kJ/day which
translates into HOW of thermal heat while asleep. The Eskimo also use a stone lamp to
light and heat the interior of the structure (producing around 300 W of thermal heat).

In the film Nannook of the North the inhabitants of the igloo and their stone lamp is
producing around 575 W of thermal energy, therefore

q = -575 W

Rewriting the Fourier Law and solving for the interior temperature 7J

T =T +

575W

Ax
A _

k

Their igloo is a half sphere of about 2.25 metres in interior height, with compacted snow
walls 0.25 m in thickness. Given its size and the fact it is a half-sphere the area of the
outside surface of the igloo is A = 31.8 m 2 . The thermal conductivity of compacted

W

snow is A: , ,, = 0.20

compacted snow

mK

Solving for the interior temperature we find that

t; = - 25 c +

575W

0.20

= - 25 C + 22.6 C
= - 2.4 C

0.25m
31.8 m 2

From a practical standpoint the interior temperature of the structure must remain below
C to prevent the structure from melting and collapsing in on its occupants. If the exterior
temperature drops to say - 50 C the Eskimo either make the walls of their igloo thicker

and cover it with a measured layer of loose snow (&

loose snow

0.05

W
mK

) or families

double up and so there is twice as much body heat with the igloo itself.

The igloo far from being a primitive structure is a simple albeit sophisticated example of
building technology. For those of you who are winter out doors type, if you ever get lost
or injured in the back country, you might build yourself a small snow home or igloo and
wait out for the rescue (see for example the SAS Handbook).

Composite Structures

In our simple example of the single glaze window pane we haven't taken into account
any heat loss through the window frame, which you may not could be significant if the
frame is made out of a thermally conductive material like aluminium.

To improve energy efficiency and to reduce heating costs it is common practice to use
windows with two or more panes or glazes of glass to significantly reduce the heat loss.
It is also common practice to make the window frames out of thermal insulating materials
such as plastic or wood instead of aluminium. Note the value of k for aluminium
compared to say wood (238:: 0.08) is a ratio of 3,000 to 1.

These considerations leads to the question, how do we take into account such composite
structures which are made with several materials with different thermal conductivity and
thickness? It is also possible (see Question 3) to determine the temperature at an
interface of materials by using the heat flux equation and solving for an indeterminate
temperature T at the interface.

Electric Analogy to Heat Transport by Conduction

The treatment of heat transport by conduction is greatly simplified by using an analogy to
current flow caused by a potential difference in a resistive material. This analogy been
electrical and thermal flow has been confirmed experimentally and therefore can be used.

In both the electrical and thermal systems

Driving Force _ .

= Resitance

Flux

Electrical Resistance:

The driving force for current flow in an electrically conductive material is the potential
difference AV . The Electrical Resistance of a material is a described by Ohm's Law,

Voltage _AV
Current I

= R

electrical

For electrical resistors in series we recall that

R Total - R \ +R 2+--^ R i

and for resistors in parallel we recall that

R

1 11 ^1

— = — + — + -" = 2j —

Total ^1 ^2 i ^

Thermal Resistance:

The driving force in thermal flow is the temperature difference AT . The Thermal
Resistance of a material is described by

AT 1

q x ~k

Ax

= R

Thermal

What this means then is that

AT 1

q x k

Ax
~A

= R

AT

Thermal

R

= q>

Thermal

We can use the basic theory of electrical resistors to also describe the thermal
conductivity of composite structures. We can then determine the value for the rate of
heat loss by first determining the total thermal resistance of the composite structure.

Example Three: Double Pane of Glass

Consider a composite structure made up of two panes of glass of area A = 1.0 m 2 each
with the same thickness, with an air space of thickness Ax = 0.005 m between the panes.
Let the interior building temperature Jj be 22 C, and the exterior building temperature T 2
be C. A schematic is given in Fig. 2: Double Pane of Glass with Air Space.

J = 21C

r 2 = OC

Fig. 2: Double Pane of Glass with Air Space

The three layers are in series and so we can use the analogy of series resistance

R Total -R l +R 2 +R 3

1

Ax l

1

+ —

k 2

Ax 2
_A_

1

+ —

K 3

Ax 3

L4_

1

"0.005m"
_1.0m 2 _

1

i

o.s

w

?4

0.0.234

^

mK
0.2256—

0.005m
1.0 m 2

mK

J 0.840.84

m^T

0.005m
1.0 m 2

Solving for the heat flow through the composite window we find

AT

R

Total

22K

0.2256

K
W

= 97.5 W

Which is considerably less than the heat loss through a single pane of glass.

Example Four: Heat Loss though an un-insulated Concrete
Wall and a Single Pane Window.

Consider a prefabricated wall 8.0 m x 5.0 m x 0.15 m thick made of un-insulated
concrete. The wall has a 1.0 m x 1.0 m window made of single paned standard glass. Let
the interior building temperature Jj be 22 C, and the exterior building temperature T 2 be
OC.

This composite system is an example of thermal resistance in parallel and as a results we
must calculate the following,

1

R

Total

R

+

1

concrete
1

R

f

Ax

N \

+

Window
1

(

V

hA) \hA

Ax-

N \

J

11 _i_ 2^

Ax, Ax^

0.8

V

mK

J

7 ( W^

(39.0m 2 ) 0.84— (l.Ow 2 )

0.15m

W W

= 208 — + 168—

K K

+ ■

J
0.005m

= 376

W_
K

The heat flux out of the building is

AT

R

Total

= (22K)376

= 8272 W

W_
K

A close look at this wall structure results in the following realization, that although the
1.0 m 2 window represents only 1/40 1 of the entire wall area, it causes 45 % of the heat
loss for the wall structure.

Example Five: Heat Loss though an Insulated Concrete Wall
and a Single Pane Window.

If we decide to leave the window as is, how much fibre glass insulated would have to be
installed to reduce the heat loss by 45% to compensate for the heat loss through the
window? Let us decide that we want to add 0.10 m of fibre glass with

W

k mre Glass = 0.035

mK

R

Insulated Wall

Concrete +&**_<*.

0.15m

f

W

\

+

0.10m

0.8
V mK j

(39.0m 2 )

f

0.035

W

\

V

mK j

(39.0m 2 )

= 0.00481 — + 0.0733 —
W W

= 0.0781 —
W

For the insulated concrete wall with insulation,

1 ' =12.81 W

r K ' i<r

iV Insulated_Wall 0733

w

The total thermal resistance for the insulated wall and single glazed window is

1 1 1

= +

D D D

Total Insulated _ Wall Window

w w

= 12.81 — + 168 —
K K

W

= 180.8 —

K

The Heat Flux for the insulated wall and single glazed window is

AT
q * = R

^ Total

= (22K)180.8 —
= 3978 W

which is about half of the heat loss through an un-insulated concrete wall with single
pane window.

Example Six: Heat Loss though an Un-insulated Concrete
Wall and a Double Pane Window.

Consider a prefabricated wall 8.0 m x 5.0 m x 0.15 m thick made of un-insulated
concrete. The wall has a 1.0 m x 1.0 m window made of double paned standard glass (to
simplify our example we shall use the result from example three for the window). Let the
interior building temperature Jj be 22 C, and the exterior building temperature T 2 be C.

This composite system is an example of thermal resistance in parallel and as a results we
must calculate the following,

1

1

R

Total

R

■ + •

1

concrete
1

R

+

Window
1

r Ax^

V

Mi

M

Ax l
f

0.2256

J

K_
W

+ 4.433

W_
K

W
0.8 —
V mK

\

(39.0m 2 )

J

0.15m

W W

= 208 — + 4.433 —

K K

+ 4.433

W_
K

= 212.433

W_

K

The heat flux for the wall with a double glazed window is

AT

-^ Total

= (22K)212.433 —
= 4673.5 W

As you see, it pays to put in better windows for it means less heat loss through the wall
structure, better energy efficiency and lower heating costs during the colder months of the
year. It is easy to convince yourself that during the warmer months of the year the better
wall system also reduces the cost to cool such a structure as well.

Example Seven: Heat Loss though an Insulated Concrete Wall
and a Double Pane Window.

It is straightforward to calculate the heat loss in the case of an insulated concrete wall
with a double glazed window pane, it is merely

AT

^ Total

= (22K)17.243 —
= 379.3 W

Which is 1/20 of that of the un-insulated concrete with a single pane window.

Homework Questions:

1 . You are by yourself and lost during a cross country skiing trip in the back country of
British Columbia. The sun is going down and the temperature is about to drop to - 40 C.
You remember learning about igloos in your BCIT physics course and you decide to
build yourself a snow structure to wait out the night and rescue. Before you retire for
the night you build a fire out side the structure far enough away that you do not melt the
igloo and bright enough to be seen a good distant away.

Design a snow structure out of loose and compact snow that will keep the inner
temperature of your refuge at no colder than -2.0 C. Proof your design by calculation.

2. As mentioned, it is common practice to make the window frames out of thermal
insulating materials such as plastic or wood, instead of aluminium. Note the value of k
for aluminium compared to say wood (238 :: 0.08) is a ratio of 3,000 to 1. Estimate the
heat loss caused by an aluminium window frame for a 1 .0 m single pane glass window.

3. Using what you know about heat flux through a composite structure derive an
expression for temperature in a wall prove that the temperature at an interface between
two materials in a wall is

_ k { Ax 2 T { + k 2 Ax { T 2

k { Ax 2 + k 2 Ax 1

The vapour barrier in a wall is a plastic film designed to keep moisture from condensing

W

on the cold outer structure of a wall. If T x = 22C and T 2 = -25C , k x = 0.035 , (fibre

mK

W

glass insulation) and & 2 =0.08 (wood exterior) and Ax 2 = 0.15m, estimate the

mK

minimum distance for Ax l to keep the interface or vapour barrier above C.

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